f(x) = 2x^2 - 4x + 3

First we need to determine f'(x)

f'(x) = 4x - 4

Now we need to find the critical values for f(x) which is f'(x)'s zero.

4x - 4 = 0

==> 4x = 4

==> x= 1

Then f'(x) has a minimmum values at x= 1

==> f(1) = 2(1) - 4(1) + 3

= 2 - 4 + 3

= 1

Then f has a minimmum value at (1, 1)

Find the minimum by taking the derivitive of the function and setting it to zero.

Use the power rule which states: d/dx [x^n] = n*x^(n-1)

f(x) = 2x^2 - 4x^1 + 3

f'(x) = 4x -4

Now set to zero

4x - 4 = 0

x = 1

minimum is at f(1) = 2 - 4 + 3 = 1 , or the point (1,1)

The function f(x) = 2x^2 - 4x + 3 has a graph that s a parabola opening upwards.

The lowest point of the graph is the value where f(x) is minimum. This can be determined by writing the equation in the vertex form. If the vertex is at (h, k) the vertex form is y = a(x - h)^2 + k. The vertex is the minimum point.

f(x) = 2x^2 - 4x + 3

= 2x^2 - 4x + 2 + 1

= 2(x^2 - 2x + 1) + 1

= 2*(x - 1)^2 + 1

The vertex is (1, 1)

The minimum value of the function f(x) = 2x^2 - 4x + 3 is 1.

f(x)=2(x^2-2x+1)+1=2(x-1)^2+1

We have 2(x-1)^2>=0

so f(x)=2(x-1)^2+1>=1

f(x) minimum=1 when x=1